The Central Dogma
(Francis Crick, 1958)
(Transcription)
DNA

(Gene/Genotype)
(Translation)
RNA

Protein
(Phenotype)
An informational process between the genetic
material (genotype) and the protein (phenotype)
Classes of RNA for
Transcription and Translation
• Informational RNA (intermediate in the process of
decoding genes into polypeptides)
– Messenger RNA (mRNA)
• Functional RNAs (never translated into proteins,
serve other roles)
– Transfer RNAs (tRNA)
• Transport amino acids to mRNA and new protein
– Ribosomal RNAs (rRNA)
• Combine with an array of proteins to form ribosomes;
platform for protein synthesis
– Small nuclear RNAs (snRNA)
• Take part in the splicing of primary transcripts in
eukaryotes
– Small cytoplasmic RNAs (scRNA)
• Direct protein traffic in eukaryotic cells
– Micro RNAs (miRNA)
• Inhibits translation and induces degradation of
complementary mRNA
RNA nucleotide sequences are
complementary to DNA molecules
New RNA is synthesized 5’ to 3’ and antiparallel to
the template
DNA template
Only one strand of the DNA acts as a
template for transcription
The template strand can be different for different genes
But….
For each gene only one strand of DNA serve as a template
Single RNA polymerase (Prokaryotes)
Core enzyme
2 ,1  and 1 ’ subunits
Holoenzyme
2 , 1 , 1 ’ subunits plus
σ subunit
Polymerizes RNA
Finds initiation sites
- 35 bases from initiation of transcription
Recognized by RNA polymerase
- 10 bases from initiation of transcription
Unwinding of DNA double helix begins here
Termination
RNA polymerase recognizes signals (sequence)
for chain termination
Releases the RNA and enzyme from the template
What would you add to an in vitro transcription system that contains an E. coli
gene for glyceraldehyde 3-phosphate dehydrogenase, an enzyme in glycolysis,
in order to get transcription th at begins from the normal transcripti on start site?
(1) RNA polymerase
(2) sigma factor
(1) ribonucleoside triphosphates with A, C, G, and U bases (rNTPs)
If you remove the TATA box and place it immediately upstream of a transcription
start site of a eukaryotic gene, and subsequently transcripti on of the mRNA is
assayed, will you still achieve transcription from the same start site?
No. The TATA box needs to be present ten nucleotides upstream of the
transcription start site to allow enough space for the proper assembly of
RNA polymerase to initiate the transcription at the start site (+1). If you
place TATA box at position 0, then transcription will now initiate at
nucleotide position +10.
Simultaneous transcription and
translation in prokaryotes
Green arrow = E. coli DNA
Red arrow = mRNA combined with ribosomes
Splicing
removes the
introns and
brings together
the coding
regions
If you were asked to isolate to tal RNA from two unknown samples and required
to identify if it were from prokaryotes or eukaryotes, what aspects regarding the
classes of RNA present will help you distinguish one from the other?
RNA from prokaryotes will contain mRNA, tRNA, rRNA. In addition to these
three types of RNA, eukaryotic sample will contain pre-mRNA, snRNA,
snoRNA, scRNA, miRNA, and siRNA.
The Central Dogma
(Francis Crick, 1958)
(Transcription)
DNA
(Gene)

(Translation)
RNA

Protein
(Phenotype)
An informational process between the genetic
material (genotype) and the protein (phenotype
Translation (protein synthesis)
Peptidyl site:
peptidyltransferase
attaches amino acid
to chain
Aminoacyl site:
new amino acid
brought in
Ribosome
moves in this
direction
Cells have adapter molecules called tRNA with a three nucleotide
sequence on one end (anticodon) that is complementary to a codon
of the genetic code.
• There are different transfer RNAs
(tRNAs) with anticodons that are
complementary to the codons for each of
the twenty amino acids.
• Each tRNA interacts with an enzyme
(aminoacyl-tRNA synthetase) that
specifically attaches the amino acid that
corresponds to its anticodon.
• For example, the tRNA to the right with
the anticodon AAG is complementary to
the UUC codon in the genetic code
(mRNA). That tRNA would carry the
amino acid phenylalanine (see genetic
code table) and only phenylalanine to the
site of protein synthesis.
• When a tRNA has its specific amino acid
attached it is said to be “charged.”
Homopolymer was then added to a test tube containing cell-free
translation system, 1 radioactively labeled amino
acid and 19 unlabeled amino acids
Proteins were isolated and checked for radioactivity
Procedure was repeated in 20 tubes, with each tube containing a
different radioactively labeled amino acid
Only one tube contained radioactively labeled protein; the amino
acid that was labeled (phenylalanine) is therefore specified by UUU
Genetic Code
Next synthesized heteropolymers
• The artificial RNA sequence would depend upon
the ratio of the two or more NDPS added
• ADP and CDP in a 1 to 5 ratio
– 1/6 probability of incorporating an A being incorporated
– 5/6 probability of incorporating a C being incorporated
• The resulting RNA molecule would be a collection
of different codons that are made-up of A and C
• The numbers of different codons in the RNA
molecule is a matter of probability
Genetic Code
ADP and CDP added in a 1 to 5 ratio AND if codon is a triplet
Possible
combinations
3A
2A : 1C
1A : 2C
3C
Probability
Possible
codons
Percent
AAA
0.4%
(1/6)2(5/6)
= 2.3%
AAC, ACA,
CAA
6.9%
(1/6)(5/6)2
= 11.6%
ACC, CAC,
CCA
(1/6)3 =
0.4%
(5/6)3 =
57.9%
CCC
(2.3 + 2.3 + 2.3)
34.8%
(11.6 + 11.6 + 11.6)
57.9%
100%
Genetic Code
The poly (AC) RNAs produced proteins containing 6 amino acids
Amino acid
Percent
Possible
codons
Proline
69%
CCC (57.9%)
2C:1A (11.6%)
1A : 2C (11.6%)
2A : 1C (2.3%)
Threonine
14%
Histidine
12%
1A : 2C (11.6%)
Asparagine
2%
2A : 1C (2.3%)
Glutamine
2%
2A : 1C (2.3%)
Lysine
1%
AAA (0.4%)
Using the table below, can you translate this nucleotide
sequence?
5’UUCGAUGCCCGGGGUCCUGAAAUUGUUCUAGA 3’
• The first step is to look for the
AUG start codon.
• Next, group the nucleotides into
a reading frame of 3
nucleotides per codons and use
the table to find the amino acid
that corresponds to each codon.
• Stop translating the mRNA
when you reach a stop codon.
• Is this what you got?
Met-Pro-Gly-Val-Leu-Lys-Leu-Phe-Stop
An mRNA has the stop codon 5ХUAA3Х.What tRNA anticodon will bind to it?
a. 5ХATT3Х
b. 5ХAUC3У
c. 5У ACU3Х
*d. none of the above
To transl ate a mRNA you require two other RNAs and they are
a. tRNA and mRNA
b. tRNA and miRNA
*c. tRNA and rRNA
d. rRNA a nd siRNA
Pathways
Gene A
Gene B
Enzyme A
Enzyme B
Substrate ------- intermediate ------ product
Most often the final product of the biochemical
pathway is something essential to life, like
amino acids, nucleotides, etc.
Pathways
Gene A
Enzyme A
Substrate ------- intermediate
Mutant Gene B
No Enzyme B
No Product
Can use mutants to work out pathways, and
identify which gene catalyzes which step
+ = growth
- = no growth
+
+
+
Minimal ornithine citrulline arginine
Wildtype
Mut 1
Mut 2
Mut 3
+
+
+
+
4
-
+
-
+
+
-
+
+
+
3
2
1
No. 1
substrate
No.
+’s

ornithine
No. 2

citrulline
No. 3

arginine